Question

How do I find the indefinite integral of
∫ (x)/(1-3x²)³

Answer 1

Set u = 1-3x², then du=-6xdx:\[\int\limits_{}^{}\frac{ xdx }{(1-3x^2)^3 }=-\frac{ 1 }{ 6 }\int\limits_{}^{}\frac{ -6xdx }{ (1-3x^2)^3 }\]Now switch to variable u:\[-\frac{ 1 }{ 6 }\int\limits_{}^{}\frac{ du }{ u^3 }=-\frac{ 1 }{ 6 }\int\limits_{}^{}u^{-3}du\]Can you do the next step?

Answer 2

Ok thank you for getting me started, I appreciate the help.

Answer 3

yw!